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FromTheMoon [43]

3 years ago

8

Seja uma carga Q=1.2×10(-8)C no vácuo, distando 40 cm de um ponto M e 50 cm de um ponto N . Determine os pontencias no ponto M e

no ponto N . ( Dados K=9×10^9 Nm^2/C )

1 answer:

Len [333]3 years ago

4 0

<span>Americano, por favor?</span>

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Line segment gj is a diameter of circle l. angle k measures (4x 6)°. circle l is inscribed with triangle g j k. line segment g j

choli [55]

The value of x in the given right triangle in a semicircle is determined as 21.

<h3>What is the measure of a triangle in a semicircle?</h3>

The triangle in a semicircle is always a right angle triangle.

From the figure shown, we can say that the triangle G J K is right triangle and m<K = 90degrees.

Given that m<K = 4x + 6, we will can use the following equation to find the value of x as shown:

4x + 6 = 90

4x = 90 - 6

4x = 84

x = 21

Thus, the value of x in the given right triangle in a semicircle is determined as 21.

Learn more about right angle here: brainly.com/question/64787

#SPJ4

8 0

2 years ago

Two people travel from Columbus to Cleveland, one by car at an average velocity of 90km/h, and one by plane at an average veloci

Flura [38]

Answer:

Explanation:

Yes , their displacement may be equal .

Suppose the displacement is AB where A is starting point and B is end point .

The car is covering the distance AB by going from A to B on straight line . On the other hand plane goes from A to C , then from C to D and then from D to B . In this way plane reaches B from A on a different path which is longer than path of the car . In the second case also displacement of plane is AB . In the second case distance covered is longer but displacement is same that is AB .

7 0

3 years ago

an object is producing a sound that has a wavelength in air of 2.69m. If the speed of sound in air is 346m/s, what is the freque

Misha Larkins [42]

Answer:

129.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 346 m / sec

wavelength ( λ ) = 2.69 m

We have to calculate Frequency ( f ) :

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 346 / 2.69 Hz

= > f = 34600 / 269 Hz

= > f = 129.74 Hz

Hence, frequency of sound is 129.74 Hz.

5 0

2 years ago

Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +19.3 km/s to −18.8 km/s o

muminat

As per above given data

initial velocity = 19.3 km/s

final velocity = - 18.8 km/s

now in order to find the change in velocity

$\Delta v = v_f - v_i$

$\Delta v = -18.8 - 19.3$

$\Delta v = -38.1 km/s$

$\Delta v = -3.81 * 10^4 m/s$

Part b)

Now we need to find acceleration

acceleration is given by formula

$a = \frac{\Delta v}{\Delta t}$

given that

$\Delta v =- 3.81 * 10^4 m/s$

$\Delta t = 2.07 years = 6.53 * 10^7 s$

now the acceleration is given as

$a = \frac{-3.81 * 10^4}{6.53 * 10^7}$

$a = - 5.84 * 10^{-4}m/s^2$

so above is the acceleration

4 0

3 years ago

An engine flywheel initially rotates counterclockwise at 6.77 rotations/s. Then, during 23.9 s, its rotation rate changes to 3.5

olganol [36]

Answer:

-2.70 rad/s²

Explanation:

Given that

ω1 = initial angular velocity of the flywheel, which is 6.77 rev/s

If we convert it to rad/s, we have

(6.77 x 2π) rad/s = 13.54π rad/s

ω2 = final angular velocity of the flywheel = -3.51 rev/s,

On converting to rad/s also, we have

(-3.51 x 2π) rad/s = 7.02π rad/s

α = average angular acceleration of the flywheel = ?

Δt = elapsed time = 23.9 s

Now, using the formula, α = (ω2 - ω1)/Δt. On substituting, we have

α = (-7.02π rad/s - 13.54π rad/s)/23.9 s

α = -20.56π rad/s / 23.9 s

α = -64.59 rad/s / 23.9 s

α = -2.70 rad/s²

Therefore, the average angular acceleration of the flywheel is -2.70 rad/s²

7 0

3 years ago